// UVa1331 Minimax Triangulation
// 刘汝佳
#include<bits/stdc++.h>
using namespace std;
#define _for(i, a, b) for (int i = (a); i < (b); ++i)
typedef long long LL;

const double eps = 1e-10;
int dcmp(double x) {
  if (fabs(x) < eps) return 0;
  return x < 0 ? -1 : 1;
}

struct Point {
  double x, y;
  Point(double x = 0, double y = 0) : x(x), y(y) {}
};

typedef Point Vector;

Vector operator+(const Vector& A, const Vector& B)
{  return Vector(A.x + B.x, A.y + B.y);}
Vector operator-(const Point& A, const Point& B)
{ return Vector(A.x - B.x, A.y - B.y); }
Vector operator*(const Vector& A, double p)
{ return Vector(A.x * p, A.y * p); }
bool operator<(const Point& a, const Point& b)
{ return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool operator==(const Point& a, const Point& b)
{ return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;}
double Dot(const Vector& A, const Vector& B)
{ return A.x * B.x + A.y * B.y; }
double Cross(const Vector& A, const Vector& B)
{ return A.x * B.y - A.y * B.x; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
bool SegmentProperIntersection(const Point& a1, const Point& a2,
                               const Point& b1, const Point& b2) {
  double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1),
         c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
  return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}

bool OnSegment(const Point& p, const Point& a1, const Point& a2) {
  return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
}

typedef vector<Point> Polygon;

int isPointInPolygon(const Point& p, const Polygon& poly) {
  int n = poly.size();
  int wn = 0;
  for (int i = 0; i < n; i++) {
    const Point& p1 = poly[i];
    const Point& p2 = poly[(i + 1) % n];
    if (p1 == p || p2 == p || OnSegment(p, p1, p2)) return -1;  // 在边界上
    int k = dcmp(Cross(p2 - p1, p - p1));
    int d1 = dcmp(p1.y - p.y);
    int d2 = dcmp(p2.y - p.y);
    if (k > 0 && d1 <= 0 && d2 > 0) wn++;
    if (k < 0 && d2 <= 0 && d1 > 0) wn--;
  }
  if (wn != 0) return 1;  // 内部
  return 0;               // 外部
}

bool isDiagonal(const Polygon& poly, int a, int b) {
  int n = poly.size();
  for (int i = 0; i < n; i++)
    if (i != a && i != b && OnSegment(poly[i], poly[a], poly[b]))
      return false;  // 中间不能有其他点
  for (int i = 0; i < n; i++)
    if (SegmentProperIntersection(poly[i], poly[(i + 1) % n], poly[a], poly[b]))
      return false;  // 不能和多边形的边规范相交
  Point midp = (poly[a] + poly[b]) * 0.5;
  return (isPointInPolygon(midp, poly) == 1);  // 整条线段在多边形内
}

const int MAXM = 50 + 2;
int M;
double D[MAXM][MAXM];
double dp(const Polygon& poly, int i, int j) {
  double& d = D[i][j];
  if (d > -1) return d;
  if (i + 1 == j) return d = 0;
  d = DBL_MAX;
  if (!(i == 0 && j == M - 1) && !isDiagonal(poly, i, j)) return d = DBL_MAX;
  _for(k, i + 1, j) {
    double m = max(dp(poly, i, k), dp(poly, k, j));
    // (i,j,k)-面积
    m = max(m, fabs(Cross(poly[i] - poly[j], poly[i] - poly[k])) / 2.0); 
    d = min(d, m);
  }
  return d;
}

// 输入一个M边形，找一个最大三角形面积最小的三角剖分，输出最大三角形的面积
double solve(const Polygon& poly) {
  _for(i, 0, M) _for(j, 0, M) D[i][j] = -1;
  return dp(poly, 0, M - 1);
}

int main() {
  int T;
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> T;
  double x, y;
  Polygon p;
  while (T--) {
    cin >> M;
    p.clear();
    _for(i, 0, M) cin >> x >> y, p.emplace_back(x, y);
    double ans = solve(p);
    printf("%.1lf\n", ans);
  }
  return 0;
}
/*
算法分析请参考: 《入门经典-第2版》 例题 9-11
*/
// 20098664 1331  Minimax Triangulation Accepted  C++11 0.000 2017-10-01 13:47:37